An Insight to References in C++

Introduction

I choose to write about references in C++ because I feel most of the people have misconceptions about references. I got this feeling because I took many C++ interviews and I seldom get correct answers about references in C++.

What is meant by references in C++? A reference is generally thought of as an aliasing of the variable it refers to. I hate the definition of references being an alias of a variable in C++. In this article, I will try to explain that there is nothing known as aliasing in C++.

Background

Both in C and in C++, there are only two ways by which a variable can be accessed, passed, or retrieved. The two ways are:

Accessing/passing variable by value

Accessing/Passing variable by address - In this case pointers will come into the picture

There is no 3rd way of accessing/passing variables. A reference variable is just another pointer variable which will take its own space in memory. The most important thing about the references is that it's a type of pointer which gets automatically dereferenced (by compiler). Hard to believe? Let's see....

A Sample C++ Code using References

Lets write a simple C++ code which will use references:

#include<iostream.h>int main()
{
int i = 10; // A simple integer variable
int &j = i; // A Reference to the variable i
j++; // Incrementing j will increment both i and j.
// check by printing values of i and j
cout<< i << j <<endl; // should print 11 11
// Now try to print the address of both variables i and j
cout<< &i << &j <<endl;
// surprisingly both print the same address and make us feel that they are
// alias to the same memory location.
// In example below we will see what is the reality
return0;
}

References are nothing but constant pointers in C++. A statement int &i = j; will be converted by the compiler to int *const i = &j; i.e. References are nothing but constant pointers. They need initialization because constants must be initialized and since the pointer is constant, they can't point to anything else. Let's take the same example of references in C++ and this time we will use the syntax that the compiler uses when it sees references.

A Sample C++ Code using References (Compiler Generated Syntax)

#include<iostream.h>int main()
{
int i = 10; // A simple integer variable
int *const j = &i; // A Reference to the variable i
(*j)++; // Incrementing j. Since reference variables are
// automatically dereferenced by compiler
// check by printing values of i and j
cout<< i << *j <<endl; // should print 11 11
// A * is appended before j because it used to be reference variable
// and it should get automatically dereferenced.
return0;
}

You must be wondering why I skipped the printing of address from the above example. This needs some explanation. Since reference variables are automatically dereferenced, what will happen to a statement like cout << &j << endl;. The compiler will convert the statement into cout << &*j << endl; because the variable gets automatically dereferenced. Now &* cancels each other. They become meaningless and cout prints the value at j which is nothing but the address of i because of the statement int *const j = &i;.

So the statement cout << &i << &j << endl; becomes cout << &i << &*j << endl; which is similar to printing the address of i in both the cases. This is the reason behind the same address being displayed while we try to print normal variables as well as reference variables.

A Sample C++ Code using Reference Cascading

Here we will try to look at a complex scenario and see how references will work in cascading. Let's follow the code below:

Here we will see if we won't depend upon the compiler to generate constant pointers in place of reference and auto dereferencing the constant pointer, we can achieve the same results.

#include<iostream.h>int main()
{
int i = 10; // A Simple Integer variable
int *const j = &i; // A Reference to the variable
// The variable j will hold the address of i
// Now we can also create a reference to reference variable.
int *const k = &*j; // A reference to a reference variable
// The variable k will also hold the address of i because j
// is a reference variable and
// it gets auto dereferenced. After & and * cancels each other
// k will hold the value of
// j which it nothing but address of i
// Similarly we can also create another reference to the reference variable k
int *const l = &*k; // A reference to a reference to a reference variable.
// The variable l will also hold address of i because k holds address of i after
// & and * cancels each other.
// so we have seen that all the reference variable will actually holds the same
// variable address.
// Now if we increment any one of them the effect will be visible on all the
// variables.
// First print original values. The reference variables will have * prefixed because
// these variables gets automatically dereferenced.
// The print should be 10,10,10,10
cout<< i << "," << *j << "," << *k << "," << *l <<endl;
// increment variable j
(*j)++;
// The print should be 11,11,11,11
cout<< i << "," << *j << "," << *k << "," << *l <<endl;
// increment variable k
(*k)++;
// The print should be 12,12,12,12
cout<< i << "," << *j << "," << *k << "," << *l <<endl;
// increment variable l
(*l)++;
// The print should be 13,13,13,13
cout << i << "," << *j << "," << *k << "," << *l <<endl;
return0;
}

A Reference Takes its Own Space in Memory

We can see this by checking the size of the class which has only reference variables. The example below proofs that a C++ reference is not an alias and takes its own space into the memory.

Conclusion

I hope that this article explains everything about C++ references. However I'd like to mention that C++ standard doesn't explain how reference behaviour should be implemented by the compiler. It's up to the compiler to decide, and most of the time it is implemented as a constant pointer.

Additional Notes to Support this Article

In the discussion forums for this article, people were having concerns that References are not constant pointers but aliases. I am writing one more example to support this fact. Look carefully at the example below:

The example using references supports the virtual mechanism, i.e. looking into the virtual pointer to get the handle to correct function pointer. The interesting thing here is how the virtual mechanism is supported by the static type which is simply an alias. Virtual mechanism is supported by dynamic information which will come into the picture only when a pointer is involved. I hope this will clarify most of the doubts.

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About the Author

I love programming and started (in 1995)even before I got some professional education on the same. Since then I've worked on IP network stack (written IPv6 stack and next generation TCP Stack), VoIP, IP Security (IKE / IPSec).
My preferred programming language is C++ and I always explore the ways to improve the A.I based systems using enhanced algorithms and data structures.

A reference is just a semantic niciety for a pointer to an object (the object may be user defined, an intrinsic value or even something not yet dreamed of) - so you can have '->' or '.' - please yourself. Other modifiers such as 'const' have no relevance except in their context. How the compiler deals with such absurdities is another question.

Sorry if I was a bit unclear. My only point was that the compiler does not handle references like const pointers. The code shows that, unlike const pointers, references can be changed after initialization.

'a' is just a name for a variable, and 'b' is just a name for that same variable.

local/filescope names refer to variables and if you define a name via reference syntax it is just the same... Another name for that variable.

consider also

class foo {
int a;
int& b;

inline foo() : b(a) {}
};

In this case a compiler can make sizeof(foo) == sizeof(a) if it wants.

Even a normal variable can be implemented as a pointer, and in some cases for some compilers it is. So the reference is no different to a normal variable.

Also, when a function is inlined, a pass by reference can avoid using an address in the assembler, while pass by pointer cannot (unless the compiler can determine that its address is never taken, of course - in which case even pass by pointer doesn't use a pointer.